Question : find inverse Laplace transform of 1/(s^2+2s-3) Solution: To find inverse Laplace transform of 1/(s^2+2s-3) Let it can be written as \[F\left(s\right)=\frac{1}{(s-1)(s+3)}\] and therefore \[F\left(s\right)=\left[\frac{1}{\left(s-1\right)}-\frac{1}{\left(s+3\right)}\right]\] Now apply Inverse Laplace transform both sides \[L^{-1}\left[F\left(s\right)\right]=L^{-1}\left[\frac{1}{(s-1)(s+3)}\right]\] \[\f(t) =e^t-e^{-3t}\] is the required inverse…
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